Question

As described in Exercise 14.41 , the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of \(60 \mathrm{kPa}\), what is the partial pressure of this substance after 60 s? (b) At what time will the partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

Short answer

(a) The partial pressure of the sulfuryl chloride after 60 seconds is approximately 16.16 kPa. (b) The time it takes for the partial pressure to decrease to one-tenth of its initial value is approximately 50.67 seconds.

Step-by-step solution

(a) Calculate the partial pressure of SO₂Cl₂ after 60 seconds

Using the first-order decay equation: \(P(t) = P_0 e^{-kt}\) Plug in the given values: \(P(60) = 60~kPa \times e^{- (4.5 \times 10^{-2}~s^{-1})(60~s)}\) Calculate the result: \(P(60) \approx 16.16 ~kPa\) The partial pressure of the sulfuryl chloride after 60 seconds is approximately 16.16 kPa.

(b) Calculate the time for the partial pressure to decrease to one-tenth of its initial value

Since we want to find the time when the partial pressure of SO₂Cl₂ is one-tenth of its initial value (60 kPa), we can set up the equation as follows: \(\frac{1}{10}(60~kPa) = 60~kPa \times e^{-(4.5 \times 10^{-2}~s^{-1})t}\) Divide both sides by the initial pressure (60 kPa): \(\frac{1}{10} = e^{-(4.5 \times 10^{-2}~s^{-1})t}\) Take the natural logarithm of both sides: \(ln(\frac{1}{10}) = -(4.5 \times 10^{-2}~s^{-1})t\) Divide by the rate constant: \(\frac{ln(\frac{1}{10})}{-(4.5 \times 10^{-2}~s^{-1})} = t\) Calculate the result: \(t \approx 50.67~s\) The time it takes for the partial pressure to decrease to one-tenth of its initial value is approximately 50.67 seconds.

Key concepts

First-order reaction

In chemistry, a first-order reaction refers to processes where the rate is directly proportional to a single reactant's concentration. This means as the concentration of the reactant decreases, the reaction rate decreases as well. The mathematical expression for a first-order reaction is given by:\[Rate = k[A]\]Here, \(k\) is the rate constant, and \([A]\) is the concentration of the reactant at any time. Notably, first-order reactions exhibit a constant half-life, which exclusively depends on the rate constant and not on the initial concentration. This makes calculations straightforward. For example, in the decomposition of sulfuryl chloride, the formulation:\[P(t) = P_0 e^{-kt}\]expresses how the pressure, representing concentration in gaseous reactions, decreases exponentially over time.

Rate constant

The rate constant, often symbolized as \(k\), is a crucial factor in kinetic studies. It provides insight into how fast a reaction proceeds. For first-order reactions, the rate constant holds the dimensions of inverse time, such as \(\mathrm{s}^{-1}\). It is independent of the concentration, depending only on the nature of the reactants and the temperature at which the reaction occurs.

For example, the rate constant for the decomposition of sulfuryl chloride at 660 K is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). This value signifies that at this specific temperature, the reaction rate progresses at a measurable and calculable pace. Changes in temperature would lead to variations in the rate constant, hence affecting the speed of the reaction.

Decomposition reaction

Decomposition reactions involve a single compound breaking down into two or more simpler substances. They are essential in various natural and industrial processes. Often requiring external energy like heat, light, or electricity to initiate, these reactions illustrate how complex molecules can degrade over time.

In our exercise, the decomposition reaction of sulfuryl chloride shows its breakdown into simpler products, described generally by:\[\text{AB} \rightarrow \text{A} + \text{B}\]During this process, sulfuryl chloride separates into sulfur dioxide and chlorine gas under certain conditions. The kinetics, particularly in first-order reactions, provide a clear understanding of the time-dependent nature of these processes. This reveals not just the speed but also the mechanism by which the decomposition occurs.